Expand and combine like terms. $(7b^5-b^2)^2=$
Answer: We can expand this expression using the "perfect square" pattern (where $P$ and $Q$ can be any monomial): $(P+Q)^2=P^2+2PQ+Q^2$ Since we have a minus sign, let's rewrite the binomial as a sum where the second term is negative, then use the pattern. $\begin{aligned} &\phantom{=}\left(7b^5-b^2\right)^2 \\\\ &=\left(7b^5+\left(-b^2\right)\right)^2 \\\\ &=(7b^5)^2+2(7b^5)(-b^2)+(-b^2)^2 \\\\ &=49b^{10}-14b^7+b^4 \end{aligned}$